Derivative of a multivariate quadraticDirection for greatest derivativeTaking derivative with respect to a vectorThe derivative of a linear operator with respect to its argumentdifficulty with the derivative of L2 normWhat's the most fundamental derivative of multivariable functions?Derivative of $f(X)=X^T$Derivative of transpose of a matrixderivative for a vectorDerivative $fracpartialpartial X^H |Tr(X X^top)|^2$Taking partial derivative of matrix
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Derivative of a multivariate quadratic
Direction for greatest derivativeTaking derivative with respect to a vectorThe derivative of a linear operator with respect to its argumentdifficulty with the derivative of L2 normWhat's the most fundamental derivative of multivariable functions?Derivative of $f(X)=X^T$Derivative of transpose of a matrixderivative for a vectorDerivative $fracpartialpartial X^H |Tr(X X^top)|^2$Taking partial derivative of matrix
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
add a comment |
$begingroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
$endgroup$
What is the derivative of vector transpose?
for example, if I want to minimize function which is written like this $frac12 x^Tx-b^Tx$ subject to $xinBbb R^n$. What will be the derivative of this function?
multivariable-calculus derivatives matrix-calculus
multivariable-calculus derivatives matrix-calculus
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
edited Mar 11 at 7:51
Rodrigo de Azevedo
13k41960
13k41960
asked Mar 10 at 14:29
ProgBProgB
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked Mar 10 at 14:29
ProgBProgB
31
asked Mar 10 at 14:29
ProgBProgB
31
31
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
ProgB is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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add a comment |
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
The function is
$$F = frac12 boldsymbolx^Tboldsymbolx-boldsymbolb^Tboldsymbolx.$$
The total derivative with respect to $boldsymbolx$ is given as
$$dF = frac12 dboldsymbolx^Tboldsymbolx+frac12 boldsymbolx^Tdboldsymbolx-boldsymbolb^Tdboldsymbolx$$
$$implies dF = dboldsymbolx^Tleft[boldsymbolx-boldsymbolb right].$$
A $dF = dboldsymbolx^TtextgradF$ comparison yields
$$textgradF = boldsymbolx-boldsymbolb.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
Your Answer
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1 Answer
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active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
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active
oldest
votes
$begingroup$
The function is
$$F = frac12 boldsymbolx^Tboldsymbolx-boldsymbolb^Tboldsymbolx.$$
The total derivative with respect to $boldsymbolx$ is given as
$$dF = frac12 dboldsymbolx^Tboldsymbolx+frac12 boldsymbolx^Tdboldsymbolx-boldsymbolb^Tdboldsymbolx$$
$$implies dF = dboldsymbolx^Tleft[boldsymbolx-boldsymbolb right].$$
A $dF = dboldsymbolx^TtextgradF$ comparison yields
$$textgradF = boldsymbolx-boldsymbolb.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
The function is
$$F = frac12 boldsymbolx^Tboldsymbolx-boldsymbolb^Tboldsymbolx.$$
The total derivative with respect to $boldsymbolx$ is given as
$$dF = frac12 dboldsymbolx^Tboldsymbolx+frac12 boldsymbolx^Tdboldsymbolx-boldsymbolb^Tdboldsymbolx$$
$$implies dF = dboldsymbolx^Tleft[boldsymbolx-boldsymbolb right].$$
A $dF = dboldsymbolx^TtextgradF$ comparison yields
$$textgradF = boldsymbolx-boldsymbolb.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
The function is
$$F = frac12 boldsymbolx^Tboldsymbolx-boldsymbolb^Tboldsymbolx.$$
The total derivative with respect to $boldsymbolx$ is given as
$$dF = frac12 dboldsymbolx^Tboldsymbolx+frac12 boldsymbolx^Tdboldsymbolx-boldsymbolb^Tdboldsymbolx$$
$$implies dF = dboldsymbolx^Tleft[boldsymbolx-boldsymbolb right].$$
A $dF = dboldsymbolx^TtextgradF$ comparison yields
$$textgradF = boldsymbolx-boldsymbolb.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
$endgroup$
The function is
$$F = frac12 boldsymbolx^Tboldsymbolx-boldsymbolb^Tboldsymbolx.$$
The total derivative with respect to $boldsymbolx$ is given as
$$dF = frac12 dboldsymbolx^Tboldsymbolx+frac12 boldsymbolx^Tdboldsymbolx-boldsymbolb^Tdboldsymbolx$$
$$implies dF = dboldsymbolx^Tleft[boldsymbolx-boldsymbolb right].$$
A $dF = dboldsymbolx^TtextgradF$ comparison yields
$$textgradF = boldsymbolx-boldsymbolb.$$
edited Mar 10 at 14:44
J.G.
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
edited Mar 10 at 14:44
J.G.
29.9k22947
edited Mar 10 at 14:44
J.G.
29.9k22947
edited Mar 10 at 14:44
J.G.
29.9k22947
29.9k22947
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
answered Mar 10 at 14:40
MachineLearnerMachineLearner
5927
5927
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
Thank you a lot for this answer, the only unclear part for me is the b transpose part how it became b
$endgroup$
– ProgB
Mar 10 at 15:13
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
$boldsymbolb^Tdboldsymbolx$ is a scalar. Hence, we can transpose it $dboldsymbolx^Tboldsymbolb$ to obtain the same scalar. Then factor $dboldsymbolx^T$.
$endgroup$
– MachineLearner
Mar 10 at 15:50
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
$begingroup$
thanks, now everything is clear
$endgroup$
– ProgB
Mar 10 at 15:59
add a comment |
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
ProgB is a new contributor. Be nice, and check out our Code of Conduct.
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